# Volume of solid from cross-sectional area

I'm a new member so please bear with me.

The diagram in the picture represents the cross sectional area of a 3D solid formed with identical circles and the solid is revolved around a vertical axis. How do you find the volume of the solid? Can the solid be transformed into a cylinder? • The answer to your first question is yes, you can find the volume from the area. My question is, do you need an analytic solution, or will a numerical solution suffice. As to your second question, imagine that the volume is a piece of clay. Can you transform it into a cylinder? Certainly. My next question is, do you understand complex variables? (I know, it seems unrelated.) – Cye Waldman Jul 9 '17 at 16:41
• I believe a numerical solution will suffice. I think 'transform' is the wrong term to use. What I mean is if the volume of the solid is equivalent to the volume of a cylinder with radius 0.6cm and height 0.6cm. – Cikguseven Jul 9 '17 at 21:14
• See my answer below. However, what you say about the cylinder is impossible because that would fill the entire space between the arrows and is larger than the volume in the question. – Cye Waldman Jul 9 '17 at 22:39

Actually, I've resolved how get an analytic solution. First of all, you only look at the positive $x$ side of the area, then divide it into an upper and lower portion. You should be able to do the lower portion by yourself, it a hemisphere and it volume is $4\pi r^3/6$. Now for the upper portion, as I showed for a similar post by someone else (it's a homework problem, right?), here is how to do the upper part.

Let's consider the problem of rotating the upper curve, a circular arc, about the $y$-axis. Let's just call the curve $y(x)$ for the moment. I think it's well known and understood that

$$V=2\pi\int x~y(x)~dx$$

The problem here is that it's not immediateley obvious how to express $y(x)$. However, if the diameter of the arc is $r$, then we can write $y(x)$ as a quarter circle centered at $(x,y)=(r,r)$ as follows

$$x=r+r\cos\theta\\ y=r+r\sin\theta\\ \theta\in[\pi,3\pi/2]$$

or, equivalently

$$x=r(1-\cos\theta)\\ y=r(1-\sin\theta)\\ \theta\in[0,\pi/2]$$

We have now parameterized $x$ and $y$, so we need to rewrite the volume expression as

$$V=2\pi\int_0^{\pi/2}x(\theta)~y(\theta)\frac{dx}{d\theta}~d\theta$$

You should be able to take it from here. If not I can extend this answer.

• Just to clarify, isn't the volume of hemisphere 2/3\pir^3? Also, shouldn't the volume of a curve revolving around the vertical axis be \pi\int_a^b x^2 dy – Cikguseven Jul 10 '17 at 2:38
• Yes on the hemisphere, both ways of calculating the other integral must be the same. Try it! – Cye Waldman Jul 10 '17 at 3:02
• May I have the numerical answer for the volume of the solid? I'm still learning integration at high school level and do not understand how you derived the equation of the arc from trignometrical functions. I thought that the equation of the quadrant/arc is based on that of a circle.[Something like (x-0.6)^2 + (y-0.6)^2 = 0.6^2] – Cikguseven Jul 10 '17 at 3:17
• If you don't understand what I did you should just go with what you know. But $x^2+y^2=r^2$. – Cye Waldman Jul 10 '17 at 3:25